The answer to problem B is clearly 1/3 and I think we both will agree here. The problem A is the same question that is asked to SB - on a given wake up event, she is asked in the moment about the probability of the coin showing heads. SO the answer in problem A is also 1/3. — PhilosophyRunner
Can you explain why the payoff tables you've come up with are unsatisfactory to you? — Pierre-Normand
The coin toss result determines the Tuesday awakening, while the Monday awakening is independent of it. — Pierre-Normand
It’s not the same because she isn’t given a randomly selected waking after 52 weeks. She’s given either one waking or two, determined by a coin toss.
The manner in which the experiment is conducted matters. — Michael
I think the Halfer position is roughly that there are only two outcomes: a single interview conducted in one sitting, and a double interview spread out over two sittings. Those outcomes are equivalent to the two possible outcomes of the coin toss. (If you have an even-numbered population to work with, you can just do away with the coin altogether.)
What is the Thirder equivalent? If there are three outcomes, they cannot be equivalent to the two outcomes of the coin toss. — Srap Tasmaner
But she is only asked a question once in the whole year. One of the wakings is randomly selected to be the one where she is asked the question. On this randomly selected waking, she is asked the question "what is the probability that this randomly selected waking shows a heads." The answer is 1/3, as per Problem A in my previous post. — PhilosophyRunner
Some researchers are going to put you to sleep. During the [experiment], they will briefly wake you up either once or twice, depending on the toss of a fair coin (Heads: once; Tails: twice). After each waking, they will put you to back to sleep with a drug that makes you forget that waking. When you [are awake], to what degree ought you believe that the outcome of the coin toss is
Heads?
Patient: Doctor, Doctor, it hurts if I do this.
Doctor: Then don't do that.
Consider the die example: When the die lands on 'six', you can't distinguish whether this outcome is from the fair die or the loaded one. — Pierre-Normand
But there are two sources of randomness in this example, the die and the coin.
Similarly for all analyses that treat SB's situation as describable with two coin flips. We only have one. — Srap Tasmaner
The halfer position comes back to individuation, as you suggested some time ago. Roughly, the claim is that "this interview" (or "this tails interview" etc) is not a proper result of the coin toss, and has no probability. What SB ought to be asking herself is "Is this my only interview or one of two?" The chances for each of those are by definition 1 in 2.
She also knows that the fact that she is awake eliminates (H,H) as a possibility. This is a classic example of "new information" that allows her to update the probabilities. With three (still equally likely) possibilities left, each has a posterior probability of 1/3. Since in only one is coin C1 currently showing Heads, the answer is 1/3. — JeffJo
I'll throw in one last consideration. I posted a variation of the experiment here.
There are three beauties; Michael, Jane, and Jill. They are put to sleep and assigned a random number from {1, 2, 3}.
If the coin lands heads then 1 is woken on Monday. If the coin lands tails then 2 is woken on Monday and 3 is woken on Tuesday.
If Michael is woken then what is his credence that the coin landed heads?
Michael's credence before the experiment is P(1) = 1/3, so if woken he ought to continue to have a credence of P(1) = 1/3 since he gains no new relevant evidence if he wakes up during the experiment. — Michael
And given that if woken he is 1 iff the coin landed heads, he ought to have a credence of P(Heads) = 1/3.
Do we accept this?
If so then the question is whether or not Sleeping Beauty's credence in the original experiment should be greater than Michael's credence in this experiment. I think it should.
Your proposed scenario certainly provides an interesting variation, but it doesn't quite correspond to the structure of the situation typically discussed in literature, the one that seems to give rise to a paradox. — Pierre-Normand
And in the "scenario most frequently discussed," there is a fourth potential outcome that halfers want to say is not a potential outcome. SB can be left asleep on Tuesday. This is an outcome in the "laboratory" space whether or not SB can observe it. It needs to be accounted for in the probability calculations, but in the "frequent discussions" in "typical literature," the halfers remove it entirely. Rather than assign it the probability it deserves and treating the knowledge that it isn't happening as "new information."In your scenario, there are four potential outcomes from the experiment, each of which is equally probable. — Pierre-Normand
And in the "scenario most frequently discussed," there is a fourth potential outcome that halfers want to say is not a potential outcome. SB can be left asleep on Tuesday. This is an outcome in the "laboratory" space whether or not SB can observe it. It needs to be accounted for in the probability calculations, but in the "frequent discussions" in "typical literature," the halfers remove it entirely. Rather than assign it the probability it deserves and treating the knowledge that it isn't happening as "new information." — JeffJo
What if the experiment ends after the Monday interview if heads, with the lab shut down and Sleeping Beauty sent home? Heads and Tuesday is as irrelevant as Heads and Friday. — Michael
Then, in the "scenario most frequently discussed," SB is misinformed about the details of the experiment. In mine, the answer is 1/3. — JeffJo
It may still be that the answer to both is 1/3, but the reasoning for the second cannot use a prior probability of Heads and Tuesday = 1/4, because the reasoning for the first cannot use a prior probability of Heads and Second Waking = 1/4.
But if the answer to the first is 1/2 then the answer to the second is 1/2. — Michael
P(Unique) = 1/3, as one-third of the experiment's awakenings are unique. — Pierre-Normand
So we have two different versions of the experiment: — Michael
This is a non sequitur. — Michael
What we can say is this:
P(Unique|Heads)=P(Heads|Unique)∗P(Unique)/P(Heads)
We know that P(Unique | Heads) = 1, P(Heads | Unique) = 1, and P(Heads) = 1/2. Therefore P(Unique) = 1/2.
Therefore P(Unique|W) = 1/2.
And if this experiment is the same as the traditional experiment then P(Heads|W) = 1/2.
I think Bayes’ theorem shows such thirder reasoning to be wrong.
P(Unique|Heads)=P(Heads|Unique)∗P(Unique)/P(Heads)
If P(Unique) = 1/3 then what do you put for the rest? — Michael
Similarly:
P(Heads|Monday)=P(Monday|Heads)∗P(Heads)P(Monday)
If P(Monday) = 2/3 then what do you put for the rest?
Previously you've been saying that P(Heads) = 1/2. — Michael
Would you not agree that this is a heads interview if and only if this is a heads experiment? If so then shouldn't one's credence that this is a heads interview equal one's credence that this is a heads experiment? — Michael
If so then the question is whether it is more rational for one's credence that this is a heads experiment to be 1/3 or for one's credence that this is a heads interview to be 1/2.
Rationality in credences depends on their application. It would be irrational to use the credence P(H) =def |{H-awakenings}| / |{awakenings}| in a context where the ratio |{H-runs}| / |{runs}| is more relevant to the goal at hand (for instance, when trying to survive encounters with lions/crocodiles or when trying to be picked up at the right exit door by Aunt Betsy) and vice versa. — Pierre-Normand
However, it's important to note that while these biconditionals are true, they do not guarantee a one-to-one correspondence between these differently individuated events. When these mappings aren't one-to-one, their probabilities need not match. Specifically, in the Sleeping Beauty problem, there is a many-to-one mapping from T-awakenings to T-runs. This is why the ratios of |{H-awakenings}| to |{awakenings}| and |{H-runs}| to |{runs}| don't match. — Pierre-Normand
I think you're confusing two different things here. If the expected return of a lottery ticket is greater than its cost it can be rational to buy it, but it's still irrational to believe that it is more likely to win. And so it can be rational to assume that the coin landed tails but still be irrational to believe that tails is more likely. — Michael
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